User:TakuyaMurata/Continuous functions on a compact space

In this section, we will undergo a throughout study of , the space of all real-valued or complex-valued functions on a compact space. The most important result in the line of this study is Ascoli's theorem and Stone-Weierstrass theorem. For simplicity, we assume functions in are real-valued. (A discussion will be given later as to why this does not diminish the generality.)

As usual, we topologizes by the norm . To say that is complete is precisely:2. LemmaThe limit of a uniformly convergent sequence of continuous functions is continuous.
Proof: Suppose is a sequence such that for some function defined on . For any , by the iterated limit theorem,

Hence, is a Banach space. (For the definition and basic results of Banach spaces, see Functional Analysis.)

2 Theorem (Ascoli)Let . Then is relatively compact if and only if

(i) Given an and , we can find a neighborhood of such that

for every and every

(ii) for every

Proof: First, assume that is relatively compact. (ii) is then obvious. For (i), let and be given. For each , by continuity, we can find a neighborhood of such that:

for every .

Since is relatively compact, contains a finite subset such that is the union of the sets of the form

over . Let be the intersection of over . Then for every , there is with , and so:

for any .

This proves (i). Next, suppose satisfies (i) and (ii). To show that is totally bounded, let be given. For each , by (i), we can find a neighborhood of such that:

for every and .

Since is compact, we can find such that is the union of over . Let

.

By (ii), is a bounded (thus totally bounded) subset of . That means that contains a finite subset such that:

It now follows: given , we can find such that:

.

Then, for each , since for some ,

In other words, . Hence, is totally bounded, or equivalently, relatively compact.

2 CorollaryLet . is uniformly convergent if and only if it is pointwise convergent and equicontinuous.

2 TheoremLet converge pointwise to . If exists and converges uniformly to , then converges uniformly to . Moreover, is differentiable and its derivative is . Proof: Let . is finite by uniform convergence. By the mean value theorem,

Thus, is equicontinuous and converges uniformly by Ascoli's thoerem (or one of its corollaries.)

2 TheoremLet be an equicontinous set of real-valued functions on . If , then there exist an such that:

Proof: Let be such that:

for all and

( isn't a typo; it is meant to simplify the computation.) Let be fixed. Then, for any ,

,

and we estimate:

where is such that . Thus,

Since we can get the same estimate for , the proof is complete.

2 Corollary (Dini's theorem)Let be a sequence such that for every . If is increasing, then .
Proof: Set . Then is decreasing and thus satisfies the hypothesis of Ascoli's theorem. Hence, admits a convergent subsequence, which converges to 0 since the subsequence converges pointwise to 0. Since is decreasing, converges as well.

2 TheoremSuppose is a metric space. Then is equicontinuous if and only if for every there exists such that

for every and with . Proof: holds vacuously. For the converse, let be given. Then for each , we can find such that

implies for every

By compactness, we find such that:

Let , and then suppose we are given with . It follows: there is a with . Since

,

we have:

for every .

The Stone-Weierstrass theorem states that polynomials are dense in C(K, R). It is however not the case that the space of polynomials in z is a dense in C(K, C). If it were, we have the equality in the below

But if K has nonempty interior.

2 Theorem (intermediate value theorem)A function is continuous if and only if

(i) If , then c is in .

(ii) If is closed for every real c.

Proof: () Obvious. () Suppose . Since the complement of , which contains x, is open, we have: in some interval U in containing x. We actually have: in U. In fact, if and , then , which implies U contains a point z such that , a contradiction. Hence, f is upper semicontinous at x. The same argument applied to shows that f is also lower semicontinous at x.

2 TheoremLet be a real-valued continuous function on an open interval. Then the following are equivalent.

(i) f is injective.

(ii) f is strictly monotonic.

(iii) f is an open mapping.

Proof: (ii) (i) is obvious. (iii) (ii): If (ii) is false, then we can assume there exists such that and . By continuity and compactness, f attains a maximum in some point x in but by hypothesis and so is a non-interior point of , falsifying (iii). If (iii) is false, then contains a x such that is not an interior point of . Since is an interval, we may assume that . It then follows from the intermediate value theorem that f is not injective.

2 Theorem (mean value theorem)Suppose is differentiable on the open interval . Then

for some
Proof: FIrst assume . By the theorem preceding this one, attains a maximum or minimum at ; say, a maximum. By definition, we can write:

as

Then since x is a maximum,

.

If ,

and letting gives that . If , then by the same argument, we find that . Thus, . For the general case, let

where .

Then . Hence, applying the first part of the proof gives: for some c. Since , c is a solution of the equation.

2 CorollaryLet be a differentiable function. If , then c is in .
Proof:Let . Then and . In other words, is increasing at a and decreasing at b. By the theorem above, g is not injective on ; i.e., for some in . It follows:

is a partition of unity, by the binomial theorem applied to . Moreover, a simple computation gives the identity:

It thus follows: for any

Since is uniformly continuous on by compactness, the theorem now follows.

2 CorollaryAny continuous function vanishing at infinity is uniformly approximated by Hermite function. (TODO: made the statement more precise and give a proof.)

Example: Let . Let be the linear span of polynomials. Then is a dense subspace of by the above theorem since

2 TheoremIf is uniformly continuous and integrable, then .
Proof: Define . That is integrable means that exists and is finite. Let be given. By uniform continuity, there is a such that

whenever .

Then there is an such that

whenever

Now, let be given. By the mean value theorem, we find such that and . Thus,

ExampleNo function is continuous only on rational points. To see this, let be a function, and let be the set of all points at which is continuous. It follows immediately from the definition of continuity that is ; i.e., it is an intersection of countably many open sets. On the other hand, is not .

2 TheoremLet be a nonempty open subset.

Proof: Let and . We assume and are finite; otherwise the inequality is trivial. Given , we can find so that an interval . By Taylor's formula,

(where )

and so:

Now, take .

2 Theorem (Whitney extension theorem)Every real-valued L-Lipschitz function on a subset of is the restriction of a L-Lipschitz function on .
Proof ([2] pg. 5): Let be a L-Lipschitz function on a subset . Define